Question

Understanding dual tangent space using the definition $T_p \to \mathbb{R}$?

Answer 1

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✔️✔️Clearly, this is not a scalar, but we need to find some object ωω such that its action makes it a scalar. The first thing I can think of that fulfills this role is the right action of ω=fω=f, a scalar function itself. However, Sean Carroll precisely states that ffshould not be thought of as a one form. Instead, we define the left action of the gradient of ff as the one form. From (1.52) we may write

dfddλ=(∂σf)θ^(σ)vμe^(μ).dfddλ=(∂σf)θ^(σ)vμe^(μ).

Now, I see how this proof could be completed and yield dfdλdfdλ--what we would attain through just the right action of 

ff--if we require θ^(σ)e^(μ)=δσμθ^(σ)e^(μ)=δμσ.

But my understanding is that this restriction is not required. I believe (but am not certain) that this would end up being circular anyway, as the orthogonality relation is precisely imposed on dual spaces, and we are trying to prove that the gradient of ff is the one form that maps 

Tp→RTp→R.

But then I don't see how this expression ends up equalling dfdλdfdλ.




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Answer 2

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